percentage

Question

a manager notices that the mean weight of the bag of grain from a recent shipment is 50 pounds with a standard deviation of 3. at least what percentage of values can the manager say will lie within 4 standard deviations of the mean?

Answer 1

I will assume a normal distribution.

Let X be the mean weight of the bag of grain.

X~normal(50, 3^2)

P(46 <= X <= 54)
= P((46 - 50) / 3 <= Z <= (54 - 50) / 3)
= P(-4/3 <= Z <=  4/3)
= 1 - P(Z < -4/3) - P(Z > 4/3)
= 1 - P(Z > 4/3) - P(Z > 4/3)
= 1 - 2 * P(Z > 4/3)
= 1 - 2 * P(Z >= 4/3)
= 1 - 2 * 0.0918
= 1 - 0.1836
= 0.8164

Hence, the manager can say that 81.64% of the values will lie within 4 standard deviations of the mean